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## Q. 3.22

The industrial process for making the ammonia used in fertilizer, explosives, and many other products is based on the reaction between nitrogen and hydrogen at high temperature and pressure:

$N_{2}(g) + 3 H_{2}(g) → 2 NH_{3}(g)$

If 18.20 kg of $NH_{3}$ is produced by a reaction mixture that initially contains 6.00 kg of $H_{2}$ and an excess of $N_{2}$, what is the percent yield of the reaction?

## Verified Solution

Collect and Organize We know that the actual yield of $NH_{3}$ is 18.20 kg. We also know that the reaction mixture initially contained 6.00 kg of $H_{2}$ and that $H_{2}$ must be the limiting reactant because the problem specifies the presence of excess $N_{2}$. We need to find the percent yield.

Analyze We need to use the mass of $H_{2}$ to calculate how much $NH_{3}$ could have been produced— the theoretical yield. We can then use the theoretical yield and the actual yield to calculate percent yield. We need the molar masses of $H_{2}$ and $NH_{3}$ and the stoichiometry of the reaction, which tells us that 2 moles of $NH_{3}$ are produced for every 3 moles of $H_{2}$ consumed.

Solve The molar masses we need are:

$\mathscr{M}_{H_{2}} = 2(1.008 g/mol) =2.016 g/mol$

$\mathscr{M}_{NH_{3}} =14.01 g/mol + 3(1.008 g/mol) = 17.03 g/mol$

We calculate the theoretical yield of $NH_{3}$:

$6.00 \sout{kg H_{2}}\times \frac{10^{3} \sout{g H_{2}}}{1 \sout{kg H_{2}}} \times \frac{1 \sout{mol H_{2}}}{2.016 \sout{g H_{2}}} \times \frac{2 \sout{mol NH_{3}}}{3 \sout{mol H_{2}}} \times \frac{17.03 \sout{g NH_{3}}}{1 \sout{mol NH_{3}}} \times \frac{1 kg NH_{3}}{10^{3} \sout{g NH_{3}}} =33.8 kg NH_{3}$

Then we divide the actual yield by this theoretical yield to determine the percent yield:

$\frac{18.20 \sout{kg NH_{3}}}{33.8 \sout{kg NH_{3}}} \times 100 \%=53.8\%$

Think About It A yield of about 54% may seem low, but it may be the best that can be achieved for a particular process. A great deal of chemical research goes into trying to improve the percent yield of industrial chemical reactions.